Powerpoint Templates Page 1 Powerpoint Templates Chapter 7 An Economic Appraisal II: NPV, AE, IRR Technique

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Chapter 7An Economic Appraisal II:NPV, AE, IRR Technique

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Net Present Value Technique

NPV=The Sum of The Present Values of All Cash Inflows – The Sum of The Present Value of All Cash Outflows

2 3 4 5 0 1

Cash Inflows

Cash Outflows

0

PV(i) CIF

PV(i) COF

NPV

NPV(i) > 0


NPV....

- Equation

- Decision Rule NPV(MARR) > 0 Accept it

NPV(MARR) = 0 Indifferent NPV(MARR) < 0 Reject it

0

0 1 2

( )(1 )

( / , ,0) ( / , ,1) ( / , , 2)

( / , , )

Nt

tt

N

CFNPV MARR

MARR

CF P F MARR CF P F MARR CF P F MARR

CF P F MARR N

Where, CFt: cash flow at time t,

MARR: minimum attractive rate of return on a project


The Steps to Make a Decision with the NPV Technique

- Step 1: Determine an MARR.

- Step 2: Estimate a project life.

- Step 3: Calculate a net cash flow(all cash inflows – all cash outlfows)

- Step 4: Calculate a net present value with an MARR.

- Step 5: Make a Decision on the Project with the NPV Derived in Step 4.


Ex 7.1] An Investment Decision on A Construction Project of A Small Power Plant with A NPVNPV

Given] Cash Flows Diagram, MARR=8%

5060

80

50

100

150

100

60

120 120

0-1-2-3-4-5-6-7-8-9-10

1 2 3 504948

……………

n

(unit: Million won)


Sol] - Step 1: MARR=8% Already Determined. - Step 2: A project life turns out be 60 years including a construction time of 10

years. - Step 3: A net cash flows are presented in the cash flow diagram. - Step 4: Calculate the net present value. (1) a present value of all the net cash flows incurred under construction.

(2) a present value of all the net cash flows incurred during the

commercialization stage of 50 years

Continued……..

A Present Value of The construction Costs

50( / ,8%,2) 60( / ,8%,3) 80( / ,8%,4) 50( / ,8%,5)

100( / ,8%,2)( / ,8%,5) 150( / ,8%,8) 100( / ,8%,9)

60( / ,8%,10) 509.85

P A P F P F P F

P A P F P F P F

P F M

120( / ,8%,50)( / ,8%,10) 680P A P F M A Present Val ue of Al l the Cash I nfl ows


- Step 5: Make an investment decision on the project with NPV(8%)

Since NPV(8%)=170.15 M >0, accept the project

Continued……..

(8%) 680 509.85 170.15NPV M

0 .5 1 .0 1 .5 2 .0

20 0

15 0

10 0

50

50

10 0

MARR(100%)

(NP

V U

nit: $M

Break_Even Interest Rate=IRR=9.78%

The Sensitivity Analysis of the NPV with A Varying Interest Rate


Net Future Value Technique

Given: Cash Flows and MARR (i)

Find: A Net Equivalent Value at the End of a Project Life

75,000

24,400 27,34055,760

01 2 3

Project Life

(unit: 000 won)


Sol] (1) A future value of all the cash flows incurred under construction

(2) A present value of all the cash flows incurred during the

commercialization stage

An Investment Decision with A NFW for Ex. 7.1

60( / ,8%,0) 100( / ,8%,1) 150( / ,8%,2)

100( / ,8%,2)( / ,8%,3) 50( / ,8%,5)

80( / ,8%,6) 60( / ,8%,7) 50( / ,8%,2)( / ,8%,8)

1,100.7

F P F P F P

F A F P F P

F P F P F A F P

M

Futue Val ue of Al l the Cash Outfl ows

120( / ,8%,50) 1,468P A M Present Val ue of Al l the Cash Fl ows

(8%) 1,468 1,100.7 367.3 ) ,accept the project.Since NFV M


- Project Balance (PB): Cash Flows Left inside A Project- Equation

PB0=PB0

PBn=PBn-1(1+MARR)+CFn

Ex 7.2] An Economic Meaning of An NPV Based on A PB

A Project Balance Concept

n 0 1 2 3 4 5

NCF (62,500) 33,982 33,726 33,205 33,135 82,013

0

1

2

3

4

5

62,500,000

62,500(1 0.15) 33,982 37,893,000

37,893(1 0.15) 33,726 9,851,000

9,851(1 0.15) 33,205 21,876,000

21,876(1 0.15) 33,135 58,292,000

58,292(1 0.15) 82,013 149,049,000

PB

PB

PB

PB

PB

PB

(unit: 000 won)


Continued…

NPV(15%) = 149,049 (P/F, 15%, 5) =74,107,0009,851

2 2.333,205

DPBP years

(unit: 000 won)

Beginning Bal.

Interest

Ending Bal.

NFV(15%)

NN 00 1 1 2 2 3 3 4 4 5 5

-62,500

-9,375

+33,982

-37,893

-37,893

-5,684

+33,726

-9,851

-9,851

-1,478

+33,205

+21,876

+21,876

+3,281

+33,135

+58,292

+58,292

+8,744

+82,013

+149,049

-62,500

-62,500


-62,500

-37,893

-9,851

21,876

58,292

160,000

140,000

120,000

100,000

80,000

60,000

40,000

20,000

0

-20,000

-40,000

-60,000

-80,000 n

PB at the End of a Project Life

Discounted Payback Peirod

PB

0 1 2 3 4 5

149,049

A Project Balance Diagram as A Function of Time


Principle: a present value of cash flows which are oriented with an equivalent amount of money of “A” over an infinite period of time.

Equation

Capitalized Equivalent

( ) ( / , , )A

CE i P A P A ii

A

P = CE(i)

1 1:

1

ni

Note P A ni i


Ex 7.3] CE

Given] A=200 M won, i= 8%, N= ∞

Sol] Calculate a CE to prepare for an annual maintenance and repair cost of a building

Continued ….

2(8%) ( / ,8%, ) 25( )

0.08CE A P A 억원

200M

P = CE(8%)=2.5B

∞


Annual Equivalent

2 3 4 5 N1

A

0 1 2 3 4 5 N

AE(i) =NPV(i)(A/P, i, N)

0

NPV(i)

0

……

……

Decision Rule - if AE(i) > 0, accept the project -if AE(i) = 0, remain indifferent -if AE(i) <, reject the project


Ex 7.4] Convert the irregular cash flows into an equivalent worth

Continued…..

2 3 4 5 6

1

A=$1.835

0 1 2 3 4 5 60

Unit:: $M

0

(15%) $6.946NPV M(15%) $6.946( / ,15%,6) $1.835AE A P M

15

3.5

59

12 108


1. Consistency of report formats: Financial managers more commonly work with annual rather than with overall costs in any number of internal and external reports. Engineering managers may be requires to submit project analyses on an annual basis for consistency and ease of use by other members of the corporation and stockholders.

2. Need for unit costs/profits: In many situations, projects must be broken into unit costs/profits for ease of comparison with alternatives. Make-or-buy and reimbursement analyses aree key examples.

3. Unequal project lives: Comparing projects with unequal service lives is complicated by the need to determine the lowest common multiple life. For the special situation of an indefinite service period and replacement with identical projects, we can avoid this complication by use of AN analysis.

The Advantages of An AE Technique


Operating Costs: to be costs which are incurred repeatedly over the life of a

project by the operation of physical plant and or equipment needed to provide servicee suck as labor and raw materials.

Capital Costs: to be costs which are incurred only one time over the life of a

project by purchasing assets to be used in production and service such as a purchase cost and sales taxes.

Capital Cost and Operating Cost


When only costs are involved, an AE cost analysis may be useful.

A profit must exceed a sum of operating and capital costs such that a project be economically viable.

Annual Equivalence Analysis

CC

OC

+

AE

Cost


Capital Recovery Cost(CR) Definition: to be an

annul equivalent of capital costs

Items of costs (1) Initial cost being the

same as the cost basis(I) (2) Salvage value(S)

CR(i) : Considering two costs

above, we obtain the following expression.

0

N

I

S

0 1 2 3 N

CR(i)

………………

iSNiPASI

NiFASNiPAIiCR

),,/)((

),,/(),,/()(


Calculate a CR(i)

- CR(i) for Mini Cooper

Unit: 000 won

Type ModelPurchase

Cost

SV at the end of year

3

Small Mini Cooper 19,800 12,078

(6%) ( )( / ,6%, )

(19,800 12,078)( / ,6%,3) 12,078(0.06) 3.61355

R I S A P N iS

P M

C

A


A Relationship between a CR(i) and a Depreciation Cost

In a practical term, a CR(i) cost consists of (1)a depreciation cost and (2) an interest on the investment cost.

I=19.8M, N = 3 years, i=6%, S=12.078M A Depreciation Method: A SL Method

19.800 12.0782.574

3D M

n Begin. Inv. Cost Interest with i= 6% PV of the interest

1 19.800 1.188 1.188(P/F,6%,1)=1.12075

2 17.226 1.03356 1.03356(P/F,6%,2)=0.919.6

3 14.652 0.87912 o.87912(P/F,6%,3)=0.73813

총액 2.77874

AE of the Interest = 2,778.74(A/P, 10%, 3)=1.03995M


Ex 7.5]

Given] I=20M, S=4M, A=4.4M, N=5 years, i= 10%

Sol] An investment decision-making with AE, and make a decision with the AE

- Method 1: first obtain the NPV and transform it into AE

- Method 2: Make a decision with CR(i)

AE-CR(i)

(10%) 20,000 4,400( / ,10%,5) 4,000( / ,10%,5)

0.83688

(10%) 836.88( / ,10%,5) 0.22076

NPV P A P F

AE A P

1

2

1 2

( ) ( )

[(20,000 4,000)( / ,10%,5) (0.10)(4,000)]

4.62076

( ) 4.400

(10%) ( ) ( )

4,620.76 4,400 0.22076

AE i CR i

A P

M

AE i M

AE AE i AE i

M


Ex 7.6] profit/machine time used – when the operating time is constant

Given] NPV=3.553M, N=3 years, i= 15%, machine time used /year: 2,000 hrs

Sol] Saving/machine time used

24,400

0

1 2 3

55,76027,340

75,000

Operating hrs

2,000hrs 2,000hrs 2,000hrs

(15%) 3,553( / ,15%,3) 1.556AE A P M

1.556 / 2,000

780 /

M hrs

hr

Savi ng/ MCH


- Def 1: ROR is the interest rate earned on the unpaid balance of an amortized loan

- Ex: A bank lend 10 million won and receives 4.021 million won each

year over the next 3 years. Then, it can be said that this bank earns 10% of ROR on the loan.

Rate of Return or Internal Rate of Return

0

1

2

3

Unit:000

n

Begin. Unrecovered

Bal.-10,000

-10,000

-6,979

-3,656

-1,000

-698

-366

Interest on the Unrec.

Bal.(10%)

Recov. Money

+4,021

+4,021

+4,021

Ending. Unrec.Bal

-10,000

-6,979

-3,656

0


- Def 2: ROR is the break-even interest rate, i*, which equates the present value of a project’s cash outflows to the present value of its cash inflows.

- Equation

ROR

* * *

**

0

31 20 * * 2 * 3

1* 1 *

( ) ( ) ( ) 0

( )(1 )

(1 ) (1 ) (1 )

(1 ) (1 )

0

Nt

tt

N NN N

NPV i PV i PV i

CFNPV i

i

CFCF CFCF

i i i

CF CF

i i

cash i nfl ows cash outfl ows


ROR=IRRInternal rate of return is the interest rate charged on the unrecovered project balance of the investment such that, when the project terminates, the unrecovered project balance will be zero

0

1

2

3

Unit:000

n

Begin. Unrecovered

Bal.-10,000

-10,000

-6,979

-3,656

-1,000

-698

-366

Interest on the Unrec.

Bal.(10%)

Recov. Money

+4,021

+4,021

+4,021

Ending. Unrec.Bal

-10,000

-6,979

-3,656

0


Simple Investment• Def: one in which the

initial cash flows are negative, and only one sign change occurs in the net cash flow series.

• Example: -100, 250,300 (-, +, +)

• i* : Only one unique i* i* becomes the IRR

Nonsimple Investment• Def: one in which more

than one sign change occurrs in the cash flow series

• Example: -100, 250,300(-, +, +,-)• i* : the real i* may exist as

many as a number of sign changes in the cash flow series.

• So, any i* can not be the IRR.

The Types of Projects


Ex 7.7]

Given] cash flows(refer to the table below)

Simple and Nonsimple Investment Project

Period (N)

Project A Project B Project C

0 -225 -270 -450

1 135 158 270

2 2,025 158 90

3 90 90

4 90 -45

5 90

6 67

Unit: 000


Sol] Identify a number of sign changes in the net cash flows

Continued….

n 0 1 2 3 4 5 6# of sign changes

S or NS

Project A

- + + 1 S

Project B

- + + + + 1 S

Project C

- + + + - + + 3 NS


Ex 7.8] understanding the IRR conept

given] I= 49.950M, Cash Inflow= 18.5M, Life= 6 years

Sol] Determine the IRR -obtain # of real root using Mathematica

IRR Concept

Plot[-49950+18500/(1+i)1+18500/(1+i)2+18500/(1+i)3

+18500/(1+i)4+18500/(1+i)5+18500/(1+i)6,{i, -1, 1}, PlotRange {-50000,300000}]

1 .0 0 .5 0 .5 1 .0

50 000

50 000

100 000

150 000

200 000

250 000

300 000

i

** * 2 * 3

18,500 18,500 18,500( ) 49,950

(1 ) (1 ) (1 )NPV i

i i i

0)1(

500,18

)1(

500,18

)1(

500,186*5*4*

iiiIRR=29%


- Check up IRR=29% with the PB concept

Prove it with a PB

nnn CFIRRPBPB )1(1

PB at “0” :

PB at “1” :

PB at “2”:

PB at “3” :

0 0 49.950PB CF M

1 49,950(1 0.29) 18,500

45.939

PB

M

2 45,939(1 0.29) 18,500

40.757

PB

M

3 40,757(1 0.29) 18,500

34.076

PB

M

PB at “4” :

PB at “5” :

PB at “6”:

5 25,458(1 0.29) 18,500

14.341

PB

M

6 45,939(1 0.29) 18,500

40.757

PB

M

4 34,076(1 0.29) 18,500

25.458

PB

M


A Decision Rule for the case in which there exists only a unique i*.

if IRR > MARR, accept the project

if IRR = MARR, remain indifferent

if IRR > MARR, reject the project

Note that this decision rule can not be applied for the case in which there exist more than one i*.


Ex 7.9] Determine the IRR of the project

Given] Cash Flow Diagrams

Calculate IRR

(a) investing project

1 2 3 4년

5,000

5

1,318.99

0 1 2

3 4 5 6 7년

90,000

180,000

90,000

80,000

120,000

150,000

(b) borrowing project

0

Unit: 000


Sol] 1) Determine the IRR of Project (a)

- Obtain a numberr of i*

- Determine the IRR

Continued…..

Plot[-90000-180000/(1+i)-90000/(1+i)2+80000/(1+i)3+ 80000/(1+i)4+120000/(1+i)5+120000/(1+i)6+150000/(1+i)7,{i,-1,1},PlotRange {-200000,500000}]

1 .0 0 .5 0 .5 1 .0

2 0 0 0 0 0

1 0 0 0 0 0

1 0 0 0 0 0

2 0 0 0 0 0

3 0 0 0 0 0

4 0 0 0 0 0

5 0 0 0 0 0

i

NPV

FindRoot[-90000-180000/(1+i)-90000/(1+i)2+80000/(1+i)3+80000/(1+i)4+120000/(1+i)5+120000/(1+i)6+150000/(1+i)70, {i,0.05}]� {i*0.10473633423827057}


1) Determine the IRR of Project (b)

- Obtain a number of i*.

- Find out the IRR

Continued……

Plot[5000-1318.99/(1+i)-1318.99/(1+i)2-1318.99/(1+i)3-1318.99/(1+i)4-1318.99/(1+i)5, {i,1,1}, PlotRange {-5000,5000}]

1 .0 0 .5 0 .5 1 .0

4000

2000

2000

4000

i

NPV

FindRoot[5000-1318.99/(1+i)-1318.99/(1+i)2-1318.99/(1+i)3-1318.99/(1+i)4-1318.99/(1+i)50, {i,0.05}]{i� 0.100001}


Ex 7.11] An Economic Analysis with a Variety of The TechniquesGiven] Cash Flows, MARR=8%Sol]

An Example Which Is Economically Analyzed with a Variety of The Appraisal Techniques

n NCF of Project A NCF of Project B NCF of Project C

0

1

2

3

4

5

6

-200,000

20,000

40,000

60,000

60,000

60,000

68,000

-200,000

80,000

60,000

60,000

40,000

-200,000

60,000

60,000

60,000

60,000

40,000

20,000

총 이익 108,000 40,000 100,000

PBP(Pref. B, C,A) 4.3 yrs 3yrs 3.3 yrs

DPBP(Pref.: B, C,A) More than 6yrs 3.9 yrs 4.05yrs

ARR* 9% 5% 8.3%

NPV(8%) 28,230 2,546 38,554

IRR 12% 8.5% 15%

Unit: 000


1. Keep the alternatives independent or mutually exclusive one another

2. Set up a common time period

3. Perform an incremental analysis if neccessay(IRR)

The Conditions to Compare Alternatives


(1) NPV - It is most widely used by companies

- Its final result is expressed in an absolute value

(2) AE - Its format is consistent with a fiscal year

- It provides a unit cost and profit - It makes us convenient to compare alternatives whose lives are

different

(3) IRR - Its final result is expressed as percentage such that managers easily

understand its meaning- It is also one of the techniques which are most widely used by

companies

The Reasons for Why NPV, AE, and IRR Techniques Are Chosen to Determine A

Preference Ordering


Ex 7.12] Determine a preference ordering of the alternatives with a variety of the techniques

Given] Cash Flow, MARR=10%, It is assumed that a capital budgett can be provided without limit

Determine a preference ordering of the alternatives

n A B C D

0

1

2

3

-

100,000

-

100,000

200,000

200,000

-

100,000

140,000

-10,000

-

100,000

50,000

-50,000

200,000

-

100,000

470,000

-

720,000

360,000

Unit: 000


- In case which the alternatives are independent( A, C, D)

Conclusion: Since all the alternatives are independent and their NPVs are greater than 0, it is better to undertake all of them.

- In case which they are mutually exclusive ( A, C, D)

Conclusion: It is required to undertake alternative “A” only because its NPV is

greater than others

A Preference Ordering with the NPV

(10%) 100,000 100,000( / ,10%,1) 200,000( / ,10%,3) 200,000( / ,10%,4)

124.640

(10%) 100,000 50,000( / ,10%,1) 50,000( / ,10%,3) 200,000( / ,10%,4)

53.950

(10%) 100,000 470,000( / ,1

A

C

D

NPV P F P F P F

M

NPV P F P F P F

M

NPV P F

0%,1) 720,000( / ,10%,3) 360,000( / ,10%,4)

2.705

P F P F

M

(10%) 124.640 (10%) 53.950 (10%) 2.705A C DNPV M NPV M NPV M


(1) Obtain the AE with a least common multiple of 6 years

Cash Flow of “A” over 6 years

n0 1

2 3

100,000

200,000 200,000

100,000

4

5 6

n0

1 2 3

100,000

140,000

4 5 6

140,000 140,000

Cash Flow of “B”

10,000

Determine a preference ordering with the AE

10,00010,000

100,000100,000


- A LCM of 6 years

Calculate the AE of “A”

n A B C D

0

1

2

3

4

5

6

-100,000

-100,000

200,000

100,000

-100,000

200,000

200,000

-100,000

140,000

-110,000

140,000

-110,000

140,000

-10,000

-100,000

50,000

-50,000

100,000

50,000

-50,000

200,000

-100,000

470,000

-720,000

260,000

470,000

-720,000

360,000

(10%) 100,000 100,000( / ,10%,1) 200,000( / ,10%,2) 100,000( / ,10%,3)

100,000( / ,10%,4) 200,000( / ,10%,5) 200,000( / ,10%,6)

218.289

ANPV P F P F P F

P F P F P F

M

(10%) (10%)( / ,10%,6)

218,289( / ,10%,6)

50.120

A AAE NPV A P

A P

M

A Comparison with the AE Technique


For “B”

For “C”

(10%) 19,008 19,008( / ,10%,2) 19,008( / ,10%, 4)

19,008{(1 ( / ,10%,2) ( / ,10%,4)}

47.700

BNPV P F P F

P F P F

M

(10%) (10%)( / ,10%,6)

47,700( / ,10%,6)

10.952

B BAE NPV A P

A P

M

(10%) 54,395 54,395( / ,10%,3)

54,395{(1 ( / ,10%,3)}

95.262

CNPV P F

P F

M

(10%) (10%)( / ,10%,6)

95,262( / ,10%,6)

21.873

C CAE NPV A P

A P

M

Continued…


For “D”

Way to Calculate the AE of the Alternatives with a single cycle of cash flows

For “ A”

(10%) 2,705 2,705( / ,10%,3)

2,705{(1 ( / ,10%,3)}

4.737

DNPV P F

P F

M

(10%) (10%)( / ,10%,6)

4,737( / ,10%,6)

1.088

D DAE NPV A P

A P

M

(10%) (10%)( / ,10%,3)

124,643( / ,10%,3)

50.120

A AAE NPV A P

A P

M

(10%) 100,000 100,000( / ,10%,1) 200,000( / ,10%,2) 200,000( / ,10%,3)

124.643ANPV P F P F P F

M

Continued…


For “B”

For “ D”

it is recommended to use the AE techniques for which the project lives are different

- When the alternatives are independent - Select the alternative with a highest AE

(10%) (10%)( / ,10%,2)

19,008( / ,10%,2)

10.952

B BAE NPV A P

A P

M

(10%) (10%)( / ,10%,3)

2,705( / 10%,3)

1

,

.088

D DAE NPV A P

A P

M

for “C”

(10%) (10%)( / ,10%,3)

54,395( / ,10%,3)

21.873

C CAE NPV A P

A P

M

Continued…


-Question: Can we determine the preference ordering of the alternatives according to the seize of the IRR?

n

0

1

IRR

NPV(10%)

Unit: 000

A1

-1,000

2,000

100%

818

A2

-5,000

7,000

40%

1,364

>

<

A Comparison with the IRR


Incremental Analysis

• If MARR=10%, it implies that the project earns a profitability of 10% is guaranteed. That is to say, the investment of 4 million won will grow up to 4.4 million won.

• We can earn 5 million won one year after once we invest 4 million won in A2. Since the IRR of 25% is greater than the MARR, it is desirable to undertake the project.

n A1 A2 (A2 – A1)

0

1

-1,000

2,000

-5,000

7,000

-4,000

5,000

IRR

NPV(10%)

100%

818

40%

1,364

25%

546

Unit: 0000


Step 1: Subtract the cash flows of Project (A) whose initial investment cost is less than the other from the cash flows of Project(B) whose initial investment cost is greater

-Step 2: Calculate the the IRRB-A of the incremental project.

Step 3: Select the project based the following decision rules.

If IRR B-A > MARR, undertake Project B If IRR B-A = MARR, remain indifferent If IRR B-A < MARR, undertake Project A.

49

The Steps of The Incremental Analysis


Ex 7.13] The Example for the Incremental Analysis with the IRR Technique

Given] Cash Flows for two projects, MARR=10%

Example

Unit: 000

n B1 B2 B2 - B1

0

1

2

3

-3,000

1,350

1,800

1,500

-12,000

4,200

6,225

6,330

-9,000

2,850

4,425

4,830

IRR 25% 17.43% 15%

If MARR = 10%, which one is the better in an economic sense?Since IRR B2-B1 >15% > 10% which is greater than MARR=10%,, it is recommended to undertake Project B2

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